By Elimu Assistant Team Five pupils A, B, C, D and E obtained the marks 53, 41, 60, 80 and 56 respectively
TOPICS: Statistics LEVEL: Form 4 Level SOURCE: Kcse 1996 Paper 1, Question 10 MARKS: 4…
TOPICS: Statistics
LEVEL: Form 4 Level
SOURCE: Kcse 1996
Paper 1, Question 10
MARKS: 4
Skills Tested:
- Statistics (mean, variance, standard deviation)
- Arithmetic (basic operations)
QUESTION
Five pupils A, B, C, D and E obtained the marks 53, 41, 60, 80 and 56 respectively. The table below shows part of the work to find the standard deviation.
(a) Complete the table (1 mark)
(b) Find the standard deviation (3 marks)
MARKING SCHEME
SOLUTION:
Question:
Five pupils A, B, C, D, and E obtained the marks 53, 41, 60, 80, and 56 respectively. The table below shows part of the work to find the standard deviation.
| Pupil | Mark x | x – x̄ | (x – x̄)² |
|---|---|---|---|
| A | 53 | -5 | 25 |
| B | 41 | -17 | 289 |
| C | 60 | 2 | 4 |
| D | 80 | 22 | 484 |
| E | 56 | -2 | 4 |
(a) Complete the table
The mean (average) of the marks is calculated as follows:
\[ x̄ = \frac{53 + 41 + 60 + 80 + 56}{5} = \frac{290}{5} = 58 \]
Completed Table:
| Pupil | Mark x | x – x̄ | (x – x̄)² |
|---|---|---|---|
| A | 53 | -5 | 25 |
| B | 41 | -17 | 289 |
| C | 60 | 2 | 4 |
| D | 80 | 22 | 484 |
| E | 56 | -2 | 4 |
(b) Find the standard deviation
To calculate the variance:
\[ \text{Variance} = \frac{\text{Sum of } (x – x̄)²}{N} = \frac{25 + 289 + 4 + 484 + 4}{5} = \frac{806}{5} = 161.2 \]
Now, we calculate the standard deviation:
\[ σ = \sqrt{161.2} \approx 12.68 \]
Final Result:
The standard deviation of the marks is approximately \( σ \approx 12.68 \).
8-4-4 Resources
